---
description: https://leetcode.com/problems/palindrome-number/
---

# 9. Palindrome Number (E)

Given an integer `x`, return `true` if `x` is palindrome integer.

An integer is a **palindrome** when it reads the same backward as forward.

* For example, `121` is a palindrome while `123` is not.

&#x20;

**Example 1:**

```
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
```

**Example 2:**

```
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
```

**Example 3:**

```
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
```

&#x20;

**Constraints:**

* `-231 <= x <= 231 - 1`

**Follow up:** Could you solve it without converting the integer to a string?

### Solution:

```
public class Solution {
    public boolean isPalindrome(int x) {
        
         if(x<0)
	      {
	    	  return false;
	      }
	      
	      //calculate how many digits it have
	      int div=1;
	      while(x/div>=10)
	      {
	    	  div=div*10;
	      }
	      
	      
	      while(x!=0)
	      {
	    	  int left=x/div;
	    	  int right=x%10;
	    	  
	    	  if(left!=right)
	    	  {
	    		  return false;
	    	  }
	    	  
	    	  x=(x%div)/10;  //remove the first digit(%div) and last digit(/10),so x is now the middle int
	    	  div=div/100;   //every time will remove two digits;
	      }
	      return true;
        
        
        
    }
}
```
